Arch324 w09 Lecture Unit6

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University of Michigan, TCAUP Structures II Slide 2/26

Architecture 324

Structures II

Reinforced Concrete by

Ultimate Strength Design

LRFD vs. ASD

Failure Modes

Flexure Equations

Analysis of Rectangular Beams Design of Rectangular Beams

Analysis of Non-rectangular Beams

Design of Non-rectangular Beams


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Allowable Stress WSD (ASD)

Actual loads used to determine stress

Allowable stress reduced by factor of safety

Ultimate Strength (LRFD)

Loads increased depending on type load

g Factors: DL=1.4 LL=1.7 WL=1.3

U=1.4DL+1.7LL

Strength reduced depending on type force

f Factors: flexure=0.9 shear=0.85 column=0.7

failureactual FSFf .).(

nu MM f

Examples:

WSD

Ultimate Strength

'1.0 cv ff

'45.0 cb ff

nu MM 9.0

nu VV 85.0

nu PP 70.0


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Strength Measurement

Compressive strength 12x6 cylinder

28 day moist cure Ultimate (failure) strength

Tensile strength 12x6 cylinder

28 day moist cure

Ultimate (failure) strength

Split cylinder test

Ca. 10% to 20% of fc

'

cf

'

tf

Photos: Source: Xb-70 (wikipedia)


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Failure Modes

No Reinforcing

Brittle failure

Reinforcing < balance

Steel yields before concrete fails

ductile failure

Reinforcing = balance

Concrete fails just as steel yields

Reinforcing > balance

Concrete fails before steel yields

Sudden failure

bd

As

yf

200min

yy

cbal

fff

870008700085.0

'

1

bal 75.0max

!h!SuddenDeatmax

Source: Polyparadigm (wikipedia)


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1

1 is a factor to account for the

non-linear shape of thecompression stress block.

f'c 10 0.85

1000 0.85

2000 0.85

3000 0.85

4000 0.855000 0.8

6000 0.75

7000 0.7

8000 0.65

9000 0.65

10000 0.65

ca 1

Image Sources: University of Michigan, Department of Architecture


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Flexure Equations actual ACI equivalentstress block stress block

bd

As



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Balance Condition

From similar triangles at balance condition:

Use equation for a. Substitute into c=a/1

Equate expressions for c:



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Rectangular Beam Analysis

Data:

Section dimensions b, h, d, (span) Steel area - As

Material propertiesfc, fy

Required: Strength (of beam) Moment - Mn

Required (by load) Moment Mu

Load capacity

1. Find= As/bd

(checkmin


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Rectangular Beam Analysis

Data:

dimensions b, h, d, (span) Steel area - As


Required:

Required Moment Mu

1. Find= As/bd

(checkmin


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Rectangular Beam Analysiscont.

2. Find a

3. Find Mn

4. Find Mu


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Slab Analysis

Data:

Section dimensions h, spantake b = 12

Steel area - As


Required:

Required Moment Mu

Maximum LL in PSF


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Slab Analysis

1. Find a

2. Find force T

3. Find moment arm z

4. Find strength moment Mn


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Slab Analysis

5. Find slab DL

6. Find Mu

7. Determine max. loading


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Rectangular Beam Design

Data: Load and Span

Material propertiesfc, fy All section dimensions b and h

Required: Steel area - As

1. Calculate the dead load and find Mu

2. d = h cover stirrup db/2 (one layer)

3. Estimate moment arm jd (or z) 0.9 d

and find As

4. Use As to find a

5. Use a to find As (repeat)

6. Choose bars for As and checkmax & min7. Check Mu


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Rectangular Slab

Design

Data: Load and Span


Required: All section dimensions h

Steel area - As

1. Calculate the dead load and

find Mu

2. Estimate moment arm

jd (or z) 0.9 d and find As

3. Use As to find a



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Rectangular Slab

Design

3. Use As to find a


5. Choose bars for As and

check As min & Asmax6. Check Mu


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Quiz 9

Can f = Mc/I be used in Ult. Strength concrete beam calculations?

(yes or no)

HINT:

WSD stress Ult. Strength stress

Source: University of Michigan, Department of ArchitectureSource: University of Michigan, Department of Architecture


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Data: Load and Span

Some section dimensions b or d


Required: Steel area - As

Beam dimensions b or d

1. Choose (e.g. 0.5max or 0.18fc/fy)2. Estimate the dead load and find Mu

3. Calculate bd2

4. Choose b and solve for db is based on form size try several to find best

5. Estimate h and correct weight and Mu

6. Find As=bd7. Choose bars for As and determine spacing

and cover. Recheck h and weight.

8. Design shear reinforcement (stirrups)

9. Check deflection, crack control, steel

development length.

8)7.14.1(

2

lwwM LLDLu

'

2

/59.01 cy

u

ffyf

Mbd

f

bdAs


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Data:

Load and Span Material propertiesfc, fy

Required:

Steel area - As

Beam dimensions b and d

1. Estimate the dead load and find Mu2. Choose (e.g. 0.5max or 0.18fc/fy)


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Rectangular Beam Design cont

3. Calculate bd2

4. Choose b and solve for d

b is based on form size.try several to find best


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Non-Rectangular Beam Analysis

Data: Section dimensions b, h, d, (span)

Steel area - As Material propertiesfc, fy

Required: Required Moment Mu (or load, or span)

1. Draw and label diagrams for section and stress1. Determing b effective (for T-beams)

2. Locate T and C (or C1 and C2)

2. Set T=C and write force equations (P=FA)1. T = As fy

2. C = 0.85 fc Ac

3. Determine the Ac required for C

4. Working from the top down, add up area tomake Ac

5. Find moment arms (z) for each block of area

6. Find Mn = Cz

7. Find Mu = fMn f=0.908. Check As min < As < As max

Source: University of Michigan, Department of Architecture


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Analysis Example

Given: fc = 3000 psi

fy = 60 ksi

As = 6 in2

Reqd: Capacity, Mu

1. Find T

2. Find C in terms of Ac

3. Set T=C and solve for Ac

Source: University of Michigan, Department of Architecture


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Example

4. Draw section and determine

areas to make Ac

5. Solve C for each area in

compression.


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Example

6. Determine moment arms to

areas, z.

7. Calculate Mn by summing

the Cz moments.

8. Find Mu = fMn


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Other Useful Tables:

Image Sources: Jack C McCormac, 1978 Design of Reinforced Concrete, Harper and Row, 1978

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Arch324 w09 Lecture Unit6

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